Find a positive $n$ for which $\mu(n)+\mu(n+1)+\mu(n+2)=3$.
As $\mu(k) \leq 1$, we must have $\mu(n)=\mu(n+1)=\mu(n+2)=1$. Moreover, it is clear that $n \equiv 1 \pmod 4$ but I don't know where to start... Also, I wrote a Python program and it turns out that $n=869$ is the smallest answer. Any hint?
The density of the square-free numbers (for which $\mu\neq 0$) is positive and equal to $\frac{6}{\pi^2}$, hence it is reasonable to expect that the set $E$ of integers $n$ such that $n,n+1$ and $n+2$ are squarefree is infinite and with density $\geq\left(\frac{6}{\pi^2}\right)^3$. Assuming a random behaviour of $\mu$ over $E$, it is reasonable to expect that about one element over eight in $E$ is such that $$\mu(n)+\mu(n+1)+\mu(n+2)=3,$$
hence the set $F$ of integers $n$ fulfulling the given constraints is expected to have a positive density around $\frac{1}{8}\left(\frac{6}{\pi^{2}}\right)^3\approx\frac{1}{36}$.
For sure, if $n\in E$ then $n\equiv 1\pmod{4}$ and $n\equiv\{1,2,3,4,5,6\}\pmod{9}$, so
$$ n\in\{1,5,\color{blue}{13},21,29,\color{red}{33}\}\pmod{36} $$
and we may notice that since $33=3\cdot 11, 34=2\cdot 17$ and $35=5\cdot 7$, we have $\color{red}{33}\in F$.
With the same sieve we may also check pretty fast that the next element of $F$ is $\color{blue}{85}$.
Moreover, the conjectured exact density of $F$ is given by
$$ \frac{1}{8}\prod_{p}\left(1-\frac{3}{p^2}\right)\approx\frac{1}{64}. $$
You found $n \equiv 1 \pmod{4}$. We also have $$ n \equiv 1,2,3,4,5,6 \pmod{9} $$ so by the chinese remainder theorem $$ n \equiv 1, 29, 21,13,5,33 \pmod{36} $$ Now, just start checking possibilities. In fact, one of the above integers ($n < 36$) works, so your program must have had an error.
